java codility training Genomic-range-query

Question

The task is:

A non-empty zero-indexed string S is given. String S consists of N characters from the set of upper-case English letters A, C, G, T.

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Answer 1

The php 100/100 solution:

function solution($S, $P, $Q) {
    $S      = str_split($S);
    $len    = count($S);
    $lep    = count($P);
    $arr    = array();
    $result = array();
    $clone  = array_fill(0, 4, 0);
    for($i = 0; $i < $len; $i++){
        $arr[$i] = $clone;
        switch($S[$i]){
            case 'A':
                $arr[$i][0] = 1;
                break;
            case 'C':
                $arr[$i][1] = 1;
                break;
            case 'G':
                $arr[$i][2] = 1;
                break;
            default:
                $arr[$i][3] = 1;
                break;
        }
    }
    for($i = 1; $i < $len; $i++){
        for($j = 0; $j < 4; $j++){
            $arr[$i][$j] += $arr[$i - 1][$j];
        }
    }
    for($i = 0; $i < $lep; $i++){
        $x = $P[$i];
        $y = $Q[$i];
        for($a = 0; $a < 4; $a++){
            $sub = 0;
            if($x - 1 >= 0){
                $sub = $arr[$x - 1][$a];
            }
            if($arr[$y][$a] - $sub > 0){
                $result[$i] = $a + 1;
                break;
            }
        }
    }
    return $result;
}

Answer 2

Java 100/100

class Solution {
public int[] solution(String S, int[] P, int[] Q) {
    int     qSize       = Q.length;
    int[]   answers     = new int[qSize];

    char[]  sequence    = S.toCharArray();
    int[][] occCount    = new int[3][sequence.length+1];

    int[] geneImpactMap = new int['G'+1];
    geneImpactMap['A'] = 0;
    geneImpactMap['C'] = 1;
    geneImpactMap['G'] = 2;

    if(sequence[0] != 'T') {
        occCount[geneImpactMap[sequence[0]]][0]++;
    }

    for(int i = 0; i < sequence.length; i++) {
        occCount[0][i+1] = occCount[0][i];
        occCount[1][i+1] = occCount[1][i];
        occCount[2][i+1] = occCount[2][i];

        if(sequence[i] != 'T') {
            occCount[geneImpactMap[sequence[i]]][i+1]++;
        }
    }

    for(int j = 0; j < qSize; j++) {
        for(int k = 0; k < 3; k++) {
            if(occCount[k][Q[j]+1] - occCount[k][P[j]] > 0) {
                answers[j] = k+1;
                break;
            }

            answers[j] = 4;
        }            
    }

    return answers;
}
} 

Answer 3

Here is my solution Using Segment Tree O(n)+O(log n)+O(M) time

public class DNAseq {


public static void main(String[] args) {
    String S="CAGCCTA";
    int[] P={2, 5, 0};
    int[] Q={4, 5, 6};
    int [] results=solution(S,P,Q);
    System.out.println(results[0]);
}

static class segmentNode{
    int l;
    int r;
    int min;
    segmentNode left;
    segmentNode right;
}



public static segmentNode buildTree(int[] arr,int l,int r){
    if(l==r){
        segmentNode n=new segmentNode();
        n.l=l;
        n.r=r;
        n.min=arr[l];
        return n;
    }
    int mid=l+(r-l)/2;
    segmentNode le=buildTree(arr,l,mid);
    segmentNode re=buildTree(arr,mid+1,r);
    segmentNode root=new segmentNode();
    root.left=le;
    root.right=re;
    root.l=le.l;
    root.r=re.r;

    root.min=Math.min(le.min,re.min);

    return root;
}

public static int getMin(segmentNode root,int l,int r){
    if(root.l>r || root.r<l){
        return Integer.MAX_VALUE;
    }
    if(root.l>=l&& root.r<=r) {
        return root.min;
    }
    return Math.min(getMin(root.left,l,r),getMin(root.right,l,r));
}
public static int[] solution(String S, int[] P, int[] Q) {
    int[] arr=new int[S.length()];
    for(int i=0;i<S.length();i++){
        switch (S.charAt(i)) {
        case 'A':
            arr[i]=1;
            break;
        case 'C':
            arr[i]=2;
            break;
        case 'G':
            arr[i]=3;
            break;
        case 'T':
            arr[i]=4;
            break;
        default:
            break;
        }
    }

    segmentNode root=buildTree(arr,0,S.length()-1);
    int[] result=new int[P.length];
    for(int i=0;i<P.length;i++){
        result[i]=getMin(root,P[i],Q[i]);
    }
    return result;
} }

Answer 4

/* 100/100 solution C++. Using prefix sums. Firstly converting chars to integer in nuc variable. Then in a bi-dimensional vector we account the occurrence in S of each nucleoside x in it's respective prefix_sum[s][x]. After we just have to find out the lower nucluoside that occurred in each interval K.

*/ . vector solution(string &S, vector &P, vector &Q) {

int n=S.size();
int m=P.size();
vector<vector<int> > prefix_sum(n+1,vector<int>(4,0));
int nuc;

//prefix occurrence sum
for (int s=0;s<n; s++) {
    nuc = S.at(s) == 'A' ? 1 : (S.at(s) == 'C' ? 2 : (S.at(s) == 'G' ? 3 : 4) );        
    for (int u=0;u<4;u++) {
        prefix_sum[s+1][u] = prefix_sum[s][u] + ((u+1)==nuc?1:0);
    }
}

//find minimal impact factor in each interval K
int lower_impact_factor;

for (int k=0;k<m;k++) {

    lower_impact_factor=4;
    for (int u=2;u>=0;u--) {
        if (prefix_sum[Q[k]+1][u] - prefix_sum[P[k]][u] != 0)
            lower_impact_factor = u+1;
    }
    P[k]=lower_impact_factor;
}

return P;

}

Answer 5

Java, 100/100, but with no cumulative/prefix sums! I stashed the last occurrence index of lower 3 nucelotides in a array "map". Later I check if the last index is between P-Q. If so it returns the nuclotide, if not found, it's the top one (T):

class Solution {

int[][] lastOccurrencesMap;

public int[] solution(String S, int[] P, int[] Q) {
    int N = S.length();
    int M = P.length;

    int[] result = new int[M];
    lastOccurrencesMap = new int[3][N];
    int lastA = -1;
    int lastC = -1;
    int lastG = -1;

    for (int i = 0; i < N; i++) {
        char c = S.charAt(i);

        if (c == 'A') {
            lastA = i;
        } else if (c == 'C') {
            lastC = i;
        } else if (c == 'G') {
            lastG = i;
        }

        lastOccurrencesMap[0][i] = lastA;
        lastOccurrencesMap[1][i] = lastC;
        lastOccurrencesMap[2][i] = lastG;
    }

    for (int i = 0; i < M; i++) {
        int startIndex = P[i];
        int endIndex = Q[i];

        int minimum = 4;
        for (int n = 0; n < 3; n++) {
            int lastOccurence = getLastNucleotideOccurrence(startIndex, endIndex, n);
            if (lastOccurence != 0) {
                minimum = n + 1; 
                break;
            }
        }

        result[i] = minimum;
    }
    return result;
}

int getLastNucleotideOccurrence(int startIndex, int endIndex, int nucleotideIndex) {
    int[] lastOccurrences = lastOccurrencesMap[nucleotideIndex];
    int endValueLastOccurenceIndex = lastOccurrences[endIndex];
    if (endValueLastOccurenceIndex >= startIndex) {
        return nucleotideIndex + 1;
    } else {
        return 0;
    }
}
}

Answer 6

I think I'm using dynamic programming. Here is my solution. Little space. Code is mot really clean, just show my idea.

class Solution {
public int[] solution(String S, int[] P, int[] Q) {
    int[] preDominator = new int[S.length()];
    int A = -1;
    int C = -1;
    int G = -1;
    int T = -1;

    for (int i = 0; i < S.length(); i++) {
        char c = S.charAt(i);
        if (c == 'A') { 
            A = i;
            preDominator[i] = -1;
        } else if (c == 'C') {
            C = i;
            preDominator[i] = A;
        } else if (c == 'G') {
            G = i;
            preDominator[i] = Math.max(A, C);
        } else {
            T = i;
            preDominator[i] = Math.max(Math.max(A, C), G);
        }
    }

    int N = preDominator.length;
    int M = Q.length;
    int[] result = new int[M];
    for (int i = 0; i < M; i++) {
        int p = P[i];
        int q = Math.min(N, Q[i]);
        for (int j = q;;) {
            if (preDominator[j] < p) {
                char c = S.charAt(j);
                if (c == 'A') {
                    result[i] = 1;
                } else if (c == 'C') {
                    result[i] = 2;
                } else if (c == 'G') {
                    result[i] = 3;
                } else {
                    result[i] = 4;
                }
                break;
            }
            j = preDominator[j];
        }
    }
    return result;
}

}